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EleoNora [17]
3 years ago
7

HELP ME RN PLEASEEEE [WILL GIVE BRAINLY TO CORRECT AND BEST EXPLAINED ANSWER]

Mathematics
2 answers:
tatuchka [14]3 years ago
8 0
D = 34 your welcome mark me please choice threee will be correct
VMariaS [17]3 years ago
6 0

Answer:

\boxed {\boxed {\sf d= \sqrt{34}}}

Step-by-step explanation:

We want to find the distance between two points, so the following formula is used.

d= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2

Where (x₁, y₁) and (x₂, y₂) are the points we are finding the distance between.

We are given the points (-2, -1) and (3,2). If we match the corresponding value and variable we see that:

  • x₁= -2
  • y₁= -1
  • x₂= 3
  • y₂= 2

Substitute the values into the formula.

d= \sqrt {(-2-3)^2+(2--1)^2

Solve the parentheses.

  • -2 -3 = -5
  • 2--1 = 2+ 1 = 3

d= \sqrt{(-5)^2+(3)^2

Solve the exponents.

  • (-5)²= -5*-5= 25
  • (3)²= 3*3=9

d= \sqrt{25+9}

Add.

d= \sqrt{34}

This radical cannot be simplified, so the distance between the two points is <u>√34</u> and <u>choice 3 </u> is correct.

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Answers:

Vertical asymptote: x = 0

Horizontal asymptote: None

Slant asymptote: (1/3)x - 4

<u>Explanation:</u>

d(x) = \frac{x^{2}-12x+20}{3x}

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Discontinuities: (terms that cancel out from numerator and denominator):

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Vertical asymptote (denominator cannot equal zero):

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<u>÷3</u>   <u>÷3 </u>

x ≠ 0

So asymptote is to be drawn at x = 0

Horizontal asymptote (evaluate degree of numerator and denominator):

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so there is NO horizontal asymptote but slant (oblique) must be calculated.

Slant (Oblique) Asymptote (divide numerator by denominator):

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  •                  -12x
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